Commutators with powers? A 'show that' question

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The discussion revolves around proving the commutator identity [AB,C] = A[B,C] + [A,C]B. Participants express uncertainty about the proof's origin, linking it to ring theory and Leibniz algebra, which are not part of their syllabus. One user attempts to derive the identity by manipulating commutators but encounters difficulties in rearranging terms. Another suggests using a test function to facilitate the proof, while others confirm the correctness of the steps taken. The conversation emphasizes the need for clarity in commutator identities and collaborative problem-solving in mathematics.
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Homework Statement



The question is 'show that the commutator [AB,C]=A[B,C]+[A,C]B'

Homework Equations



I'm not sure, a search for a proof gave the names 'ring theory' and 'Leibniz algebra', but further searching hasn't provided a proof so far and it seems it is just accepted as a standard identity. Ring theory isn't on my syllabus and I haven't heard of it before today :|

The Attempt at a Solution



I tried applying the arbitrary commutators and I got ABC-CAB, and now I'm stuck lol as I can't arrange it to fit the above. Any help would be greatly appreciated, as I'm most likely going about it the wrong way, any identities for commutators with multiplied operators within them would be useful to know :P

Thanks :)
 
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Introduce a test function f:

[AB,C]f = ABCf - CABf = A([B,C]- CB)f + ([A,C]-AC)f

and work from there. Then just drop the test function f.
 
Thanks for the tip but I think I managed to do it without the test function (I think :P)

Could somebody please go over my workings to confirm if they are correct or not?

Expand:
[AB,C] = ABC - CAB

Sub in a factor of (ACB-ACB):
[AB,C] = ABC + (ACB - ACB) - CAB

Re-arrange:
(ABC - ACB) + (ACB - CAB)

Take out common factors:

A(BC - CB) + (AC - CA)B

Use commutator identities below:
A[B,C] = A(BC - CB)
[A,C]B = (AC - CA)B

Leading to A[B,C] + [A,C]B with any luck :P

Cheers :)
 

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